Is there NO solution to this linear system of 3 equations, $3$ unknowns? I have the following linear system: 
$$\begin{align}
&x + y  + 2z + 2 = 0 \\
&3x - y + 14z -6 = 0 \\
&x + 2y +5 = 0
\end{align}$$
I immediately noticed that there was no $z$ term in the last equation and thus determined that I will end with $0z = \text{some number}$ and therefore, came to the conclusion that there is no solution to the linear system. To see if I was right, I checked with echelon form and that also suggested that there was no solution. 
$$\begin{align}
&L_1: x + y  + 2z + 2 = 0 \\
&L_2: 3x - y + 14z -6 = 0 \\
&L_3: x + 2y +5 = 0 \\
\end{align}$$
Then $-3L_1 + L_2 \rightarrow L_2$
$$\begin{align}
&x + y + 2z + 2 = 0 \\
&0x + 3y - 42z - 18 = 0 \\
&x + 2y + 0z + 5 = 0 
\end{align}$$
Then $-L_1 + L_3 \rightarrow L_3$
$$\begin{align}
&x + y + 2z + 2 = 0 \\
&0x + 3y - 42z - 18 = 0 \\
&0x + 1y + 0z + 3 = 0 
\end{align}$$
Then $-L_2/3 + L_3 \rightarrow L_3$
$$\begin{align}
&x + y + 2z + 2 = 0 \\
&0x + 3y - 42z - 18 = 0 \\
&0x + 0y + 0z + 3 = 0 
\end{align}$$ 
Firstly, is my answer correct? If so, then can I make the same conclusions for other linear systems that are similar? 
 A: Your first idea is off, because, for example the system $$x+y+z=1, x+2y=1, x+y+2z=1$$ still has a solution  even though $z$ is absent from the second equation.
On the other hand, your manipulation of equations (you call it echelon form) is one of possible correct methods of finding solutions/proving their absence.
A: First of all, you "jumped to" an erroneous conclusion, based on inspection. In a system of equations, one or more variables may fail to be present in one or more equations.
A most extreme example would be the three equations in three unknowns:
$$\begin{align}
&x - 1 = 0 \\ 
&y - 1= 0 \\
&z - 1 = 0 
\end{align}$$
From which we can "read off" the unique solution: $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1\end{pmatrix}$
There would be a problem (and no solution would exist) if you had the following (say, reduced) linear system of equations:
$$\begin{align}
&x + y  + 2z + 2 = 0 \\
&3x - y + 14z -6 = 0 \\
&0x + 0y + 0z + 5 = 0
\end{align}$$
Note that in the above system, we have the absurd equation $5 = 0$: such a system is called inconsistent, and clearly, no solution exits.
Finally, if you end up with an equation of all "zeros": $0x + 0y + 0z = 0$, then infinitely many solutions exist, and depending on how many such equations exist in your system, you might have a system with one parameter, or two, which then, while an infinite number of solutions exist, there would constraints which limit exactly which solutions are valid; i.e., the parameter(s) would define a "family" of infinitely many solutions.

Now, back to your good idea to "check out" your initial conclusion:
In your last elementary row operation, note that you didn't operate on the $-42 z - 18$ of $L_2$:
We go from:
$$\begin{align}
&x + y + 2z + 2 = 0 \\
&0x + 3y - 42z - 18 = 0 \\
&0x + 1y + 0z + 3 = 0 
\end{align}$$
Then applying, correctly, $-L_2/3 + L_3 \rightarrow L_3$
$$\begin{align}
&x + y + 2z + 2 = 0 \\
&0x + 3y - 42z - 18 = 0 \\
&0x + 0y + 14z + 9 = 0 
\end{align}$$ 
Now, you'll see that a unique solution exists: $14 z = -9 \implies $
To solve the system at this point, you might want to write your equations as follows:
$$\begin{align}
&x + y + 2z = -2 \\
&0x + 3y - 42z =18 \\
&0x + 0y + 14z = -9 
\end{align}$$ 
